首页 > 代码库 > BZOJ 1927 星际竞速(费用流)
BZOJ 1927 星际竞速(费用流)
考虑费用流,题目要求走n个点都走完且恰好一次,显然流量的限制为n。
建立源点s和汇点t,并把每个星球拆成两个点i和i‘,分别表示已到达该点和经过该点。
对于能力爆发,建边(s,i‘,1,w). 对应高速航行,建边(s,i,1,0), (i,j‘,1,w).
因为每个点必须走一次且只能走一次。建边(i‘,t,1,0).
其实就是类似最小路径覆盖的建图方法。
# include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # include <set> # include <cmath> # include <algorithm> using namespace std; # define lowbit(x) ((x)&(-x)) # define pi 3.1415926535 # define eps 1e-9 # define MOD 10000 # define INF 1000000000 # define mem(a,b) memset(a,b,sizeof(a)) # define FOR(i,a,n) for(int i=a; i<=n; ++i) # define FO(i,a,n) for(int i=a; i<n; ++i) # define bug puts("H"); # define lch p<<1,l,mid # define rch p<<1|1,mid+1,r # define mp make_pair # define pb push_back typedef pair<int,int> PII; typedef vector<int> VI; # pragma comment(linker, "/STACK:1024000000,1024000000") typedef long long LL; int Scan() { int res=0, flag=0; char ch; if((ch=getchar())==‘-‘) flag=1; else if(ch>=‘0‘&&ch<=‘9‘) res=ch-‘0‘; while((ch=getchar())>=‘0‘&&ch<=‘9‘) res=res*10+(ch-‘0‘); return flag?-res:res; } void Out(int a) { if(a<0) {putchar(‘-‘); a=-a;} if(a>=10) Out(a/10); putchar(a%10+‘0‘); } const int N=1605; //Code begin... struct Edge{ int to, next, cap, flow, cost; Edge(int _to=0, int _next=0, int _cap=0, int _flow=0, int _cost=0): to(_to), next(_next), cap(_cap), flow(_flow), cost(_cost){} }edge[40005]; struct ZKW_MinCostMaxFlow{ int head[N], tot, cur[N], dis[N], ss, tt, n, min_cost, max_flow; bool vis[N]; void init(){tot=0; mem(head,-1);} void addedge(int u, int v, int cap, int cost){ edge[tot]=Edge(v,head[u],cap,0,cost); head[u]=tot++; edge[tot]=Edge(u,head[v],0,0,-cost); head[v]=tot++; } int aug(int u, int flow){ if (u==tt) return flow; vis[u]=true; for (int i=cur[u]; i!=-1; i=edge[i].next) { int v=edge[i].to; if (edge[i].cap>edge[i].flow&&!vis[v]&&dis[u]==dis[v]+edge[i].cost) { int tmp=aug(v,min(flow,edge[i].cap-edge[i].flow)); edge[i].flow+=tmp; edge[i^1].flow-=tmp; cur[u]=i; if (tmp) return tmp; } } return 0; } bool modify_label(){ int d=INF; FO(u,0,n) if (vis[u]) for (int i=head[u]; i!=-1; i=edge[i].next) { int v=edge[i].to; if (edge[i].cap>edge[i].flow&&!vis[v]) d=min(d,dis[v]+edge[i].cost-dis[u]); } if (d==INF) return false; FO(i,0,n) if (vis[i]) vis[i]=false, dis[i]+=d; return true; } PII mincostmaxflow(int start, int end, int nn){ ss=start, tt=end, n=nn; min_cost=max_flow=0; FO(i,0,n) dis[i]=0; while (1) { FO(i,0,n) cur[i]=head[i]; while (1) { FO(i,0,n) vis[i]=false; int tmp=aug(ss,INF); if (tmp==0) break; max_flow+=tmp; min_cost+=tmp*dis[ss]; } if (!modify_label()) break; } return mp(min_cost,max_flow); } }solve; int main () { int n, m, s, t, x, u, v; scanf("%d%d",&n,&m); s=0; t=2*n+1; solve.init(); FOR(i,1,n) scanf("%d",&x), solve.addedge(s,i,1,0), solve.addedge(s,n+i,1,x), solve.addedge(n+i,t,1,0); FOR(i,1,m) { scanf("%d%d%d",&u,&v,&x); if (u>v) swap(u,v); solve.addedge(u,n+v,1,x); } printf("%d\n",solve.mincostmaxflow(s,t,t+1).first); return 0; }
BZOJ 1927 星际竞速(费用流)
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